Answer
The events are not independent.
$P(female)\ne P(female~|~5+~activities)$ and $P(5+~activities)\ne P(5+~activities~|~female).$
Work Step by Step
The sample space: 400 teens. So, N(S) = 400.
N(female) = 200, N(5+ activities) = 109 and N(female and 5+ activities) = 71. So:
$P(female)=\frac{N(female)}{N(S)}=\frac{200}{400}=0.5$. (Classical Method, page 259.)
$P(5+~activities)=\frac{N(5+~activities)}{N(S)}=\frac{109}{400}=0.2725$. (Classical Method, page 259.)
Now:
$P(female~|~5+~activities)=\frac{N(female~and~5+~activities)}{N(5+~activities)}=\frac{71}{109}\approx0.6514$. (Conditional Rule, page 288.)
$P(5+~activities~|~female)=\frac{N(female~and~5+~activities)}{N(female)}=\frac{71}{200}=0.355$. (Conditional Rule, page 288.)
$P(female)\ne P(female~|~5+~activities)$ and $P(5+~activities)\ne P(5+~activities~|~female).$
The events are not independent. See definition, page 292.